Q26 of 52 Page 1

Prove that the curves y2=4x and x2=4y divide the area of the square bound by x=0, x=4, y=4 and y=0 into three equal parts.

Or


Using Integration, find the area of the triangle whose vertices are (2,3), (3,5) and (4,4).


y2 = 4x



The area of the square bounded by x=0, x=4, y=4 and y=0 will be 4×4=16 square units.


Let f: y2=4x and g: x2=4y


Let I1 be the area bounded by the curve f and y-axis and I2 be the area bounded by the curve g and x-axis and I3 be the area bound in between f and g-curve.


Then, I1+I2+I3=16 –(1)



Hence, -(2)



Hence, –(3)


Substituting values of (2) and (3) in (1), we get


or


Hence, and therefore it divides the square in 3 equal parts.


Or



The area of the triangle will be equal to


∆ABC= (area under line segment AB) + (area under line segment BC)- (area under line segment AC)


Let line AB, BC and AC be l, m and n respectively.


Equation of l, by two-point form, will be or y=2x-1.


Equation of m, by two-point form, will be or y=8-x.


Equation of n, by two-point form, will be or .



Hence,





Hence, the area of triangle whose vertices are (2,3), (3,5) and (4,4) is square units.


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